并行wilcox.test使用group_by和summary

Question

必须有一种R-ly方式wilcox.test使用group_by并行调用多个观察.我已经花了很多时间阅读这篇文章,但仍然无法弄清楚是否wilcox.test有这样的工作.下面的示例数据和代码,使用magrittr管道和summarize().

library(dplyr)
library(magrittr)

# create a data frame where x is the dependent variable, id1 is a category variable (here with five levels), and id2 is a binary category variable used for the two-sample wilcoxon test
df <- data.frame(x=abs(rnorm(50)),id1=rep(1:5,10), id2=rep(1:2,25))

# make sure piping and grouping are called correctly, with "sum" function as a well-behaving example function 
df %>% group_by(id1) %>% summarise(s=sum(x))
df %>% group_by(id1,id2) %>% summarise(s=sum(x))

# make sure wilcox.test is called correctly 
wilcox.test(x~id2, data=df, paired=FALSE)$p.value

# yet, cannot call wilcox.test within pipe with summarise (regardless of group_by). Expected output is five p-values (one for each level of id1)
df %>% group_by(id1) %>% summarise(w=wilcox.test(x~id2, data=., paired=FALSE)$p.value) 
df %>% summarise(wilcox.test(x~id2, data=., paired=FALSE))

# even specifying formula argument by name doesn't help
df %>% group_by(id1) %>% summarise(w=wilcox.test(formula=x~id2, data=., paired=FALSE)$p.value)

越野车调用产生此错误:

Error in wilcox.test.formula(c(1.09057358373486, 
    2.28465932554436, 0.885617572657959,  : 'formula' missing or incorrect

谢谢你的帮助; 我希望它对有类似问题的其他人也有所帮助.

Answer 1

您可以使用do函数轻松完成任务(在加载dplyr库后调用？do).使用您的数据,链将如下所示:

df <- data.frame(x=abs(rnorm(50)),id1=rep(1:5,10), id2=rep(1:2,25))
df <- tbl_df(df)
res <- df %>% group_by(id1) %>% 
       do(w = wilcox.test(x~id2, data=., paired=FALSE)) %>% 
       summarise(id1, Wilcox = w$p.value)

产量

res
Source: local data frame [5 x 2]

    id1    Wilcox
  (int)     (dbl)
1     1 0.6904762
2     2 0.4206349
3     3 1.0000000
4     4 0.6904762
5     5 1.0000000

注意我在group_by和summarize之间添加了do函数. 我希望它有所帮助.

Answer 2

您可以使用基本 R 来执行此操作（尽管结果是一个繁琐的列表）：

by(df, df$id1, function(x) { wilcox.test(x~id2, data=x, paired=FALSE)$p.value })

或使用 dplyr：

ddply(df, .(id1), function(x) { wilcox.test(x~id2, data=x, paired=FALSE)$p.value })

  id1        V1
1   1 0.3095238
2   2 1.0000000
3   3 0.8412698
4   4 0.6904762
5   5 0.3095238